A rigid and insulated tank of `3 m^3` volume is divided into two compartments. One compartment of volume of `2 m^(3)` contains an ideal gas at 0.8314 MPa and 400 K and while the second compartment of volume `1m^(3)` contains the same gas at 8.314 MPa and 500 K. If the partition between the two compartments is ruptured, the final temperature of the gas is:

Options

A. 420 K
B. 450 K
C. 480 K
D. 500 K

Explanation

Mole of the gas in the first comparement , `n _(1) = ( P _(1) V_(1))/( RT _(1)) =- ( 0. 8314 xx 10 ^(6) xx 2)/(8. 314 xx 400) = 500`
Similarly, `n _(2) = 2000,` The tank is rigid and insultated hence `w =0 and q =0` therefore, `Delta U =0`
Let `T _(f) and P _(f)` denote the final temperature and pressure respectively.
`Delta U = n _(1) C _( v, m) [ T _(t) - T_(1)] + n _(2) C _( v, m ) [ T _(1) - T _(2) ]= 0,500 (T _(f) - 400) + 2000 ( T _(f) - 500) = 0 T _(f) = 48 0 K`